Sphericl Coordintes This is the coordinte system tht is most nturl to use - for obvious resons (e.g. NWP etc.). λ longitude (λ increses towrd est) ltitude ( increses towrd north) z rdil coordinte, locl verticl (z increses in locl "up" dirn) î, ĵ, re unit vectors pointing est, north nd up, respectively. u, v, w re sphericl velocity components (ltitudinl, longitudinl, nd verticl) v uî + vĵ + w [Note: sometimes use sme symbols u, v, w for Crtesin components. Should be cler from context whether u, v, w re sphericl or Crtesin.] Let's relte u, v, w to rte of chnge of prcel's sphericl coords. Intuitively, u should depend on rte of chnge of longitude λ. So consider prcel tht trvels estwrd distnce in smll time δt: rcos x (out of the pge) r rcos rcos uî From the figure we hve: rcos We know tht where we ssume tht ~r.
u lim δt 0 lim δt δt 0 rcos cos Dλ -- δt Similrly, for the N-S component in the sphericl system, we d intuit tht it depends on the ltitude, δy vĵ r δ r From our figure we hve, rδ δy. δy v lim δt 0 lim δt δt 0 r δ r D -- D -- δt nd for the verticl velocity w Dr - D ( + z) Dz - In order to use the conventionl dx, dy, nd dz we must use the following trnsformtions: Dx cosdλ, Dy rd, DzDr With these definitions, our velocities re u Dx Dy -, v -, w Dz - Don t be fooled into thinking tht this is Crtesin system however (even though it is mde to look like one with these trnsformtions). The system remins sphericl s the unit vectors, which do not remin constnt, re function of sphericl positions. Thus Du - D ( uî + vĵ + w ) î Du - ĵ Dv - Dw -- u Dî v Dĵ D + + + + + w-
We re going to evlute the chnge in the unit vectors î, ĵ, nd leve for homework problem 2.4. (styrofom blls nd stick pins Voodoo Meteorology) Hve students demonstrte tht the unit vector î depends only on longitude î î( x), therefore Dî 0 î ---- u î v î 0 + + + w---- î t x y z 0 Wht is î x? (use r~) Let s tke slice thru the erth nd perpendiculr to the xis of rottion, zoom cm î( x) î( x + ) cos î( x) cos 90- î( x + ) î( x) δî From figure, s well s number of previous exmples we know cos δî î( x + ) î( x) î where δî is to î nd directed towrd the xis of rottion. We cn decompose δî into components: ĵ î into pge δî δî cos δî sinĵ
From our figure then, δî δî sinĵ δî cos But we wnt î x î lim -- sinĵ cos sinĵ cos x 0 lim -- 0 cos 1 -- tnĵ -- Dî Of course we wnt u î, thus x Dî u î x u -- ( tnĵ ) OK, now for the chnge in the unit vector ĵ Drw our pictures & ply with our stickpins nd Michels $1.49 styrofom globe-surrogtes: For this cse we note tht ĵ chnges both ltitudinlly nd longitudinlly - so one more degree of Dĵ freedom thn the i component. How does this ply out when describing? Dĵ 0 ---- u v w 0 + + + ---- t x y z So now we need sphericl reltionships for both x, nd y... Off we go, s in Holton but with little extr info...
A δθ C 90 cos ĵ( x + ) 90 ĵ( x) B λ δθ ĵ( x) λ + From the digrm we know tht ĵ( x + ) ĵ( x) wht direction does point? Hint δθ is good indictor from the top of the digrm. δθ ĵ( x) ĵ( x + ) Thus points due west (nd is to ĵ ). Wht is the sclr mgnitude or length of? δθ ĵ So wht is δθ? From the big figure bove, AB δθ This leds to the next question... AB? Looking t tringle ABC, we get Thus, AB cos --- cot sin
δθ --- 1 -- tn ĵ δθ cot 1 -- tn Note tht points due west so î 1 ā - î tn lim x 0 lim -- 0 tn Now we need y... ĵ( y + δy) δ 90 δ δ δy ĵ( y) ĵ( y + δy) ĵ( y) Where points down nd perpendiculr to ĵ, i.e. in the - direction, nd δy δ, nd ĵ δ, thus lim y δy 0 --- δ 1 -- δy δ y -- Thus Dĵ u j + v j x y u -- v tnî -- See problem Holton 2.13 for D eqution (problem #2.4). Recll,
Du - D ( uî + vĵ + w ) î Du - ĵ Dv - Dw -- u Dî v Dĵ D + + + + + w- grouping ll of our terms, î Du - uv wu tn + - ĵ Dv - ---- wv + - Dw + + tn + -- u 2 u 2 ---- v 2 ----